# The Dirac Delta Function Potential

A delta function potential, $V(x)=V_0\delta(x-a)\!$, is a special case of the symmetric finite square well; it is the limit in which the width of the well goes to zero and the depth of the well goes to infinity, while the product of the height and depth remains constant. For such a potential, the wave function is still continuous across the potential, ie. $x=a\!$. However, the first derivative of the wave function is not.

For a particle subject to an attractive delta function potential, $V(x)=-V_0\delta(x),\!$ the Schrödinger equation is

$-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x).$

For $x \neq \!0$ the potential term vanishes, and all that is left is

$\frac{d^2\psi(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi(x)=0.$

One or more bound states may exist, for which $E<0\!$, and $\psi(x) \!$ vanishes at $x = \pm \infty \!$. The bound state solutions are given by

$\psi_{I}(x) = Ae^{kx},\, x < 0 \!$
$\psi_{II}(x) = Be^{-kx},\, x > 0 \!$

where $k=\frac{\sqrt{-2mE}}{\hbar}.$

The first boundary condition, the continuity of $\psi(x) \!$ at $x = 0 \!$, yields $A = B \!$.

The second boundary condition, the discontinuity of $\frac{d\psi(x)}{dx} \!$ at $x = 0 \!$, can be obtained by integrating the Schrödinger equation from $-\epsilon \!$ to $\epsilon \!$ and then letting $\epsilon \rightarrow 0 \!$

Integrating the whole equation across the potential gives

$-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon} \frac{d^2 \psi(x)}{dx^2}\,dx-V_0\int_{-\epsilon}^{+\epsilon} \delta(x)\psi(x)\,dx=\int_{-\epsilon}^{\epsilon} E \psi(x)\,dx$

In the limit $\epsilon \rightarrow 0 \!$, we have

$-\frac{\hbar^2}{2m}\left[\frac{d \psi_{II}(0)}{dx}-\frac{d \psi_{I}(0)}{dx}\right]-V_0\psi(0)=0,$

which yields the relation, $k = \frac{mV_0}{\hbar^2}. \!$.

Since we defined $k = \frac{\sqrt{-2mE}}{\hbar}, \!$, we have $\frac{\sqrt{-2mE}}{\hbar} = \frac{mV_0}{\hbar^2} \!$. Then, the energy is $E = -\frac{mV_0^2}{2\hbar^2} \!$

Finally, we normalize $\psi(x) \!$:

$\int_{-\infty}^{\infty} |\psi(x)|^2\,dx=2|B|^2\int_{0}^{\infty} e^{-2kx}\,dx=\frac{|B|^2}{k}=1$

so,

$B=\sqrt{k}=\frac{\sqrt{mV_{0}}}{\hbar}$

Evidently, the delta function well, regardless of its "strength" $V_{0} \!$, has one bound state:

$\psi(x)=\frac{\sqrt{mV_{0}}}{\hbar}e^{\frac{-mV_{0}|x|}{\hbar^{2}}}$

For a delta potential of the form $V_0\delta(x-a)\!$, we may apply a similar procedure to the one employed above to show that the discontinuity in the derivative of the wave function is

$\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}=\frac{2m V_0}{\hbar^2}\psi(a).$

## Problem

(Double delta function potential)

Consider a double delta function potential, $V(x)=-g\delta(x-a)-g\delta(x+a).\!$ Prove that a ground state with even parity always exists. By means of a sketch, show that a first excited state also exists for a sufficiently large potential strength $g\!$ and sufficiently large well separation $2a.\!$